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	92,768 reputation	bio	website		visits	member for	2 years, 3 months
578167 badges		location	Berlin, Germany		seen	34 mins ago

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Oct 9	comment	Formula that takes on all integers There have been two further downvotes. Could the downvoters please explain their downvotes? Please see the FAQ: "If you see misinformation, vote it down. Add comments indicating what, specifically, is wrong." Also, please note that this is an answer to a previous version of the question, as explained in my comment under the question. You can see the previous versions of the question by clicking on the link "edited ..." underneath the question.
Oct 9	comment	Pseudo Inverse Solution for Linear Equation System Using the SVD @user1551: Thanks :-)
Oct 9	comment	Spanning trees in a ladder graph @Alex: I've added an alternative solution of the recurrence to my answer.
Oct 9	revised	Spanning trees in a ladder graph added solution with matrix diagonalization
Oct 8	answered	Given a random graph $G_{n,p}$, how to get the expectation of number of components with $k$ vertices and $k$ edges?
Oct 8	answered	Systems of polynomial equations
Oct 8	revised	transforming vector potential with a coordinate rotation formatting
Oct 8	comment	Deleting any digit yields a prime… is there a name for this? That's not surprising if you look at my estimates. Solutions with few digits are rare; most of the solutions have around $10$ strings of repeated digits.
Oct 8	revised	Reference: Representation Theory spelling
Oct 8	comment	Deleting any digit yields a prime… is there a name for this? Note that my estimate predicts $0.64$ solutions with repeated last digit with $332$ digits even if the original number is prime. Dropping the condition that the original number is prime removes a factor $p_m$ and increases the expected number of solutions to $489$. For most of these the number of strings of repeated digits is around the average value of $10$; the expected number of solutions with $332$ digits, repeated last digit and only $2$ strings of repeated digits is about $0.32$.
Oct 8	comment	Tile $\mathbb{R}^n$ with Primitive Cuboids @t.b.: I see, thanks, sorry.
Oct 8	comment	Is there a clever solution to this elementary probability/combinatorics problem? +1 -- very nice :-) This also immediately yields the generalized version that the probability for $k$ balls is $2/(k+1)$.
Oct 7	comment	Tile $\mathbb{R}^n$ with Primitive Cuboids You can delete the question yourself (using the "delete" link underneath it). Also, if you've found a solution, you could write it down and accept it instead of deleting the question.
Oct 7	comment	converting integrand limits from Cartesian to spherical @Tyler: Yes, I'm sorry, I overlooked that. There's a $\sin \phi$ missing in the integrand, which comes from the Jacobian determinant (as does the factor $r^2$).
Oct 7	comment	Formula that takes on all integers Please see my comment under the question.
Oct 7	comment	Formula that takes on all integers I find this quite unacceptable. You posted a question; I pointed out that the assertion it contained was false; we engaged in a long exchange of comments, during which you kept changing and deleting your comments, forcing me to do the same; you made several confusing errors in your comments which I patiently pointed out to you; and when we had finally succeeded in clarifying your question, I wrote an answer answering it correctly. Your reaction to all this effort on my part is to change the question without marking it as changed and downvoting my correct answer to the original question?!
Oct 7	answered	Formula that takes on all integers
Oct 7	comment	Expressing the number of non-zero rows of a binary matrix as a polynomial @Ragib: You could post that as an answer so the question won't remain unanswered, since it's likely that noone will have anything to add to that.
Oct 7	revised	Expressing the number of non-zero rows of a binary matrix as a polynomial formatting
Oct 7	comment	Formula that takes on all integers Do you want to prove this for particular values of $a$,$b$,$m$? It's clearly not true for arbitrary values, since $z$ is divisible by $\gcd(a,b,m)$.